Java 第一个数字是第二个数字的倍数吗

我正在编写一个程序，要求用户输入两个整数，总共十次

接下来，程序需要评估第一个整数是否是第二个整数的倍数

如果第一个是第二个的倍数，则程序应打印为真，如果不是，则应打印为假

这是我的密码：

     public static void main(String[] args) {

         Scanner input = new Scanner(System.in);

         int counter = 0; // Initializes the counter

         System.out.printf("Enter first integer: "); // asks the user for the first integer
         int number1 = input.nextInt(); // stores users input for the first integer

         System.out.printf("Enter second integer: "); // asks the user for the second integer
         int number2 = input.nextInt(); // stores the users input for the second integer

         while (counter <= 10) // starts a loop that goes to 10
         {
             if (number1 & number2 == 0) // checks to see if number1 is a multiple of number2
                 System.out.print("true"); // if so then print out "true"
             else
                 System.out.print("false"); // otherwise print out "false"
         }

     } // end class

我的代码在某个地方被破解了。是否有人可以帮助我，或者至少为我指出正确的方向？

&是按位逻辑AND函数。我很确定那不会做你想让它做的事。似乎你想要模运算符，%

例如，使用数字1%数字2。而不是1号和2号

     while (counter <= 10) // starts a loop that goes to 10
     {
         if (number1 % number2 == 0) // checks to see if number1 is a multiple of number2
         System.out.print("true"); // if so then print out "true"

         else
         System.out.print("false"); // otherwise print out "false"
     }

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您需要将两个输入读取10次。并测试number1是否是number2的倍数。差不多

public static void main(String str[]) throws IOException {
    Scanner input = new Scanner(System.in);

    for (int counter = 0; counter < 10; counter++) {
        System.out.printf("Enter first integer for counter %d: ", counter);
        int number1 = input.nextInt();

        System.out.printf("Enter second integer for counter %d: ", counter);
        int number2 = input.nextInt();

        // Since you want to print true if number1 is a multiple of number2.
        System.out.println(number1 % number2 == 0);
    }
}

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使用%而不是&来检查number1是否是number2的倍数；然而，现在我需要程序循环十次并打印出真或假。